집단 1개, t검정

P(a < \hat\mu < b) = P\bigg(~ {a-\mu\over\sigma/\sqrt{n}} < {\overline{X}-\mu \over\sigma/\sqrt{n}} < {b-\mu\over\sigma/\sqrt{n}}~\bigg)

{\sigma\over\sqrt{n}} \cdot \sqrt{{(n-1)s^2\over\sigma^2} \cdot {1 \over n-1}} ~=~ {\sigma\over\sqrt{n}} ~\cdot~ {s \over \sigma} = {s \over \sqrt{n}}

\begin{align*} &=~P\bigg(~ {a-\mu\over s/\sqrt{n}} < {\overline{X}-\mu \over s/\sqrt{n}} < {b-\mu\over s/\sqrt{n}}~\bigg) \\[25pt] &=~P\bigg( -t_{\alpha/2}~(n-1) < {\overline{X}-\mu \over s/\sqrt{n}} < t_{\alpha/2}~(n-1)~\bigg) \\[25pt] &=~P\bigg(~ \overline{X}-t_{\alpha/2}(n-1)~ {s\over\sqrt{n}} < \mu < \overline{X}+t_{\alpha/2}(n-1)~ {s\over\sqrt{n}} ~\bigg) \end{align*}

\hat\mu = \overline{X}. \kern{15pt} X_i \sim N(\mu, \sigma^2). \kern{15pt} \text{i.i.d} \\[15pt] \text{confidence interval} = \overline{X} \pm t_{\alpha/2}(n-1)~ \cdot{s\over\sqrt{n}}